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Why is $\displaystyle \int \dfrac{dy}{dx} dx = y + c$, but for example $\displaystyle \int \dfrac{dy}{dx} y dx = \dfrac{1}{2} y^2 + c$ instead of $\dfrac{1}{2} y^3 + c$?

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3 Answers

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Using and abusing the mathematical notation as sometimes is done when dealing with differential equations, what you really have here is

$$\int\frac{dy}{\color{red}{dx}}\color{red}{dx}=\int 1\cdot dy=y+C\;,\;\;\text{since}\;\;\frac{d}{dy}(y+C)=1$$

OTOH,

$$\int\frac{dy}{\color{red}{dx}}y\,\color{red}{dx}=\int y\,dy=\frac{y^2}2+C\;,\;\;\text{since}\;\frac d{dy}\left(\frac12y^2+C\right)=y$$

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Let me integrate it out:
$$\int y'ydx=y^2-\int yy'dx,$$
by integrating by parts. Hence $\int yy'dx=(1/2)y^2.$
Hope this helps.
And feel free to ask if something above is unclear.

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I have another question and i hope someone can clarify it:

How do you find out what this is: $\int \frac{dy}{dx}$

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